Fuchs-Sondheimer Resistivity model for Nanowires

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Avocado
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I calculated FS Model using Mathcad and I got that this part
Mathcad rendering of an integral from 0 to π over dθ of sin(θ)·cos(θ)² times (1 − (1 − p)·[e^(−w/(2λ·cos(θ)·cos(φ(w)))) / (1 − p·e^(−w/(2λ·cos(θ)·cos(φ(w)))))]).


increases with decreasing width $\w$.
This make the resistivity increases with decreasing width $\w$.
This contradict the result of this model.

Has anyone ever come across with this before?

the original paper: https://journals.aps.org/prb/abstract/10.1103/PhysRevB.61.14215
complete equation:
Two-line equation for resistivity ratio rho0/rho as a triple integral over y, x and phi with nested integrals over theta, containing exp and arctan terms with w, h and lambda.
 
Attachments
  • Mathcad rendering of an integral from 0 to π over dθ of sin(θ)·cos(θ)² times (1 − (1 − p)·[e^(−w/(2λ·cos(θ)·cos(φ(w)))) / (1 − p·e^(−w/(2λ·cos(θ)·cos(φ(w)))))]).
    upload_2019-1-11_19-41-34.png
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  • Two-line equation for resistivity ratio rho0/rho as a triple integral over y, x and phi with nested integrals over theta, containing exp and arctan terms with w, h and lambda.
    vat07.png
    17.5 KB · Views: 1,169
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I don’t understand. The resistivity does increase with decreasing width. The paper you link to says this explicitly in the abstract:
The abstract said:
We find that the resistivity increases as wire width decreases in a manner which is dependent on the mean grain size and cannot be explained adequately by either model alone.
 
TeethWhitener said:
I don’t understand. The resistivity does increase with decreasing width. The paper you link to says this explicitly in the abstract:

yes, the paper does say so.

When I calculate it, the result is the opposite.

This term below,
Equation: integral from 0 to pi of sin(theta) times cos(theta) squared times (1 - (1 - p) times (exp(-w / (2 lambda cos(theta) cos(phi(w)))) / (1 - p exp(-w / (2 lambda cos(theta) cos(phi(w)))))) d theta.

increases with dicreasing width. This make the resistivity increases with decreasing width.

I calculated it with Mathcad.
The integration is difficult, so I cannot check it by hand calculation.
What I mean is, if there are some mistakes, the mistakes might be from that term.
 
Attachments
  • Equation: integral from 0 to pi of sin(theta) times cos(theta) squared times (1 - (1 - p) times (exp(-w / (2 lambda cos(theta) cos(phi(w)))) / (1 - p exp(-w / (2 lambda cos(theta) cos(phi(w)))))) d theta.
    upload_2019-1-11_19-41-34-png.png
    5.1 KB · Views: 804
I still don’t understand. You, the paper, and Mathcad all say the same thing: that resistivity increases with decreasing width. Where is the problem?
 
Apologize, what I meant is resistivity decreases with decreasing width.
This is the opposite of the model prediction.

This term below,
Mathcad rendering of an integral from 0 to π over dθ of sin(θ)·cos(θ)² times [1 − (1 − p)·(e^(−w/(2λ·cos(θ)·cos(φ(w)))) / (1 − p·e^(−w/(2λ·cos(θ)·cos(φ(w))))))].

increases with dicreasing width. This make the resistivity decreases with decreasing width.I finish calculating it with MATLAB to double check.
It shows the same with the Mathcad calculation.
Small width has smaller resistivity.

Side-by-side comparison of a MATLAB command window and a MathCad worksheet computing resistivity rho_PS(w) against width w from 65 to 95 nm, with a matching plot of the curve.
 
Attachments
  • Side-by-side comparison of a MATLAB command window and a MathCad worksheet computing resistivity rho_PS(w) against width w from 65 to 95 nm, with a matching plot of the curve.
    upload_2019-1-12_16-27-3.png
    25.4 KB · Views: 804
  • Mathcad rendering of an integral from 0 to π over dθ of sin(θ)·cos(θ)² times [1 − (1 − p)·(e^(−w/(2λ·cos(θ)·cos(φ(w)))) / (1 − p·e^(−w/(2λ·cos(θ)·cos(φ(w))))))].
    upload_2019-1-11_19-41-34-png-png.png
    5.1 KB · Views: 766
It's a bit dry to me. In your last screenshot, are the plots the ratio of ##\frac{\rho _0}{\rho}##? What is ##\rho _0##? I'm guessing it's the resistivity at a particular width. If that ratio goes to ##0## as ##w## goes to ##0## as you seem to show in your plots, then it would mean that the resistivity increases when ##w## decreases, which is inline with the paper. So there would be no problem.
 
fluidistic said:
It's a bit dry to me. In your last screenshot, are the plots the ratio of ##\frac{\rho _0}{\rho}##? What is ##\rho _0##? I'm guessing it's the resistivity at a particular width. If that ratio goes to ##0## as ##w## goes to ##0## as you seem to show in your plots, then it would mean that the resistivity increases when ##w## decreases, which is inline with the paper. So there would be no problem.

The plot is ##\rho(w)## vs. ##\w##.
##\rho _0## is the bulk resistivity, considered a constant.
 
Avocado said:
The plot is ##\rho(w)## vs. ##\w##.
##\rho _0## is the bulk resistivity, considered a constant.
I see, this is puzzling. Which value of p did you choose (1/2?).
 
fluidistic said:
I see, this is puzzling. Which value of p did you choose (1/2?).

Yes, I chose p=0.5 in order to recreate the plot on the paper.
 
I am a bit puzzled and this is not a topic I am very knowledgeable about, but why are you allowed to set [itex]\phi=arctan\frac{w}{h}[/itex]?
Is this not supposed to be an integration over the whole azimuthal angle?
 
Cthugha said:
I am a bit puzzled and this is not a topic I am very knowledgeable about, but why are you allowed to set [itex]\phi=arctan\frac{w}{h}[/itex]?
Is this not supposed to be an integration over the whole azimuthal angle?

Yes, you are right.
I am trying other formula now.
The calculation take long time though. Still not sure about the result.

Mathematica notebook screenshot defining parameters h=20 nm, w=25 nm, p=0.5, lambda=40 nm, rho0=2.44e-6 ohm-cm, integral expressions for A(w) and B(w), Ratio_FS(w)=1-A(w)-B(w), rho_FS(w)=rho0/Ratio_FS(w), with a red-bordered plotting-failed message saying to replace complex values and NaNs by real numbers.
 
Attachments
  • Mathematica notebook screenshot defining parameters h=20 nm, w=25 nm, p=0.5, lambda=40 nm, rho0=2.44e-6 ohm-cm, integral expressions for A(w) and B(w), Ratio_FS(w)=1-A(w)-B(w), rho_FS(w)=rho0/Ratio_FS(w), with a red-bordered plotting-failed message saying to replace complex values and NaNs by real numbers.
    medium?v=1.png
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